# Question #11cbb

##### 1 Answer

#### Answer:

#### Explanation:

Start by writing a balanced chemical equation that describes the combustion of a hydrocarbon

#"C"_ x"H"_ (y(g)) + (x + y/4)"O"_ (2(g)) -> x"CO"_ (2(g)) + y/2"H"_ 2"O"_((g))#

Your job here will be to determine the value of

The idea here is that when two gases are kept under **the same conditions for pressure and temperature**, the ratio that exists between their *moles* is **equivalent** to the ratio that exists between their *volumes*.

You can prove this by using the **ideal gas law** equation, which looks like this

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Now, let's say that the conditions for pressure and temperature are **total number of moles** of gas present on the reactants' side, you can say that

#P * V_1 = n_1 * RT -># the reactants' side

Likewise, if you take **total number of moles** of gas present on the products' side, you can say that

#P * V_2 = n_2 * RT -># the products' side

Divide these two equations to get

#(color(red)(cancel(color(black)(P))) * V_1)/(color(red)(cancel(color(black)(P))) * V_2) = (n_1 * color(red)(cancel(color(black)(RT))))/(n_2 * color(red)(cancel(color(black)(RT))))#

which gets you

#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_1/n_2 = V_1/V_2)color(white)(a/a)|))) -># themole ratiois equivalent to thevolume ratio

So, the total number of moles present on the reactants' side will be

#n_1 = 1 + (x + y/4)#

This is the case because **mole** of **moles** of oxygen gas.

The total number of moles present on the products' side will be

#n_2 = x + y/2#

Here, **moles** of carbon dioxide and **moles** of water are produced by the reaction.

This means that you have

#(1 + x + y/4)/(x + y/2) = (600 color(red)(cancel(color(black)("mL"))))/(700color(red)(cancel(color(black)("mL"))))#

which is equivalent to

#(1 + x + y/4)/(x + y/2) = 6/7#

At this point, all you have to do is solve this for

#1 + x + y/4 = 6 implies x = 5 - y/4#

You will have

#5 - y/4 + y/2 = 7#

#y/4 = 2 implies y = 8#

The value of

#x = 5 - 8/4 = 3#

Therefore, the **molecular formula** of this unknown hydrocarbon is

#color(green)(|bar(ul(color(white)(a/a)"C"_3"H"_8color(white)(a/a)|))) -># propane